Let be a collection of open sets, and be their union. If , then there is an with . Since is open, there is an open rectangle containing . So is open.
Let and be open, and . If , then there are open rectangles (resp. ) containing and contained in (resp. ). Since the intersection of two open rectangles is an open rectangle (Why?), we have ; so is open. The assertion about finitely many sets follows by induction.
The intersection of the open intervals is the set containing only , and so the intersection of even countably many open sets is not necessarily open.
If , then let be the open rectangle centered at with sides of length . If , then
and so . This proves that is open.
The interior of is the set ; the exterior is ; and the boundary is the set .
The interior of is the empty set ; the exterior is ; and the boundary is the set .
The interior of is the empty set ; the exterior is the empty set ; and the boundary is the set .
In each case, the proofs are straightforward and omitted.
To do the construction, first make a list of all the rational numbers in the interval [0, 1]. Then make a list of all the quarters, sixteenths, etc. of the unit sqare. For example, could be made by listing all pairs (a, b) of integers with positive, non-negative, , in increasing order of , and amongst those with same value of in increasing lexicographical order; then simply eliminate those pairs for which there is an earlier pair with the same value of . Similarly, one could make by listing first the quarters, then the sixteenths, etc. with an obvious lexicographical order amongst the quarters, sixteenths, etc. Now, traverse the list : for each portion of the square, choose the point such that is in the portion, both and are in the list , neither has yet been used, and such that the latter occurring (in ) of them is earliest possible, and amongst such the other one is the earliest possible.
To show that this works, it suffices to show that every point in the square is in the boundary of . To show this, choose any open rectangle containing . If it is , let . Let be chosen so that . Then there is some portion of the square in which is entirely contained within the rectangle and containing . Since this part of the square contains an element of the set A and elements not in A (anything in the portion with the same x-coordinate works), it follows that is in the boundary of .
Clearly, the interior of is itself since it is a union of open sets; also the exterior of clearly contains as . Since the boundary is the complement of the union of the interior and the exterior, it suffices to show that nothing in is in the exterior of . Suppose is in the exterior of . Let be an open interval containing and disjoint from . Let be a rational number in contained in . Then there is a which contains , which is a contradiction.
Suppose . Since is open, there is an open interval containing and disjoint from . Now contains a non-empty open subinterval of and this is necessarily disjoint from . But every non-empty open subinterval of contains rational numbers, and contains all rational numbers in , which is a contradiction.
Suppose is compact. Let be the open cover consisting of rectangles for all positive integers . Since is compact, there is a finite subcover . If , then and so is bounded.
To show that is closed, it suffices its complement is open. Suppose is not in . Then the collection where is an open cover of . Let be a finite subcover. Let . Then is an open neighborhood of which is disjoint from . So the complement of is open, i.e. is closed.
Such an is in the exterior of , and so there is an open rectangle containing and disjoint from . Let . This was chosen so that is entirely contained within the open rectangle. Clearly, this means that no can be , which shows the assertion.
For each , choose to be as in part (a). Then is an open cover of . Let be a finite subcover, and let . Then, by the triangle inequality, we know that satisfies the assertion.
A counterexample: is the x-axis and is the graph of the exponential function.
Let be as in Problem 1-21 (b) applied with and . Let . It is straightforward to verify that is bounded and closed; so is compact. Finally, is also true.